\(a=\lim\limits_{x\rightarrow3}\frac{\left(x-3\right)\left(2x+3\right)}{\left(x-3\right)\left(x^3+3x^2+9x\right)}=\lim\limits_{x\rightarrow3}\frac{2x+3}{x^3+3x^2+9x}=\frac{2.3+3}{3^3+2.3^2+9.3}=...\)
\(b=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^4+x^2+2x^3+2x+2\right)}=\frac{1+1}{1+1+2+2+2}=...\)
\(c=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)^2\left(4x^3+3x^2+2x+1\right)}{\left(x-1\right)^2\left(x^2+x+2\right)}=\frac{4+3+2+1}{1+1+2}=...\)
\(d=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1+1+1+1+1}{1+1+1}=...\)
\(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=Lim_{x\rightarrow3}\frac{x\left(x^3-3^3\right)}{\left(x-3\right)\left(2x+3\right)}\)
\(=Lim_{x\rightarrow3}\frac{x\left(x-3\right)\left(x^2+3x+9\right)}{\left(x-3\right)\left(2x+3\right)}=Lim_{x\rightarrow3}\frac{x\left(x^2+3x+9\right)}{2x+3}\)
\(=\frac{3\left(3^2+3.3+9\right)}{3.2+3}=\frac{3\left(9+9+9\right)}{9}=9\)
Vậy \(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=9\)
\(Lim_{x\rightarrow1}\frac{x^5+x^3-2}{x^2-1}=Lim_{x\rightarrow1}\frac{x^5-x^4+x^4-x^3+2x^3-2x^2+2x^2-2x+2x-2}{\left(x-1\right)\left(x+1\right)}\)
\(=Lim_{x\rightarrow1}\frac{\left(x^4+x^3+2x^2+2x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=Lim_{x\rightarrow1}\frac{x^4+x^3+2x^2+2x+2}{x+1}\)
\(=\frac{1+1+2+2+2}{1+1}=\frac{4.2}{2}=4\)
Vậy \(Lim_{x\rightarrow1}\frac{x^5+x^3-2}{x^2-1}=4\)
\(Lim_{x\rightarrow1}\frac{4x^5-5x^4+1}{\left(x-1\right)\left(x^3+x-2\right)}=Lim_{x\rightarrow1}\frac{4x^5-4x^4-x^4+1}{\left(x-1\right)\left[2\left(x-1\right)\left(x+1\right)+2\left(x-1\right)\right]}\)
\(=Lim_{x\rightarrow1}\frac{4x^4\left(x-1\right)-\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}{2\left(x-1\right)^2\left(x+2\right)}\)
\(=Lim_{x\rightarrow1}\frac{\left(x-1\right)\left(4x^4-x^3-x^2-x-1\right)}{\left(x-1\right)^2\left(2x+4\right)}\)
\(=Lim_{x\rightarrow1}\frac{\left(x-1\right)\left[x^3\left(x-1\right)+x^2\left(x^2-1\right)+x\left(x^3-1\right)+\left(x^4-1\right)\right]}{\left(x-1\right)^2\left(2x+4\right)}\)
\(=Lim_{x\rightarrow1}\frac{\left(x-1\right)^2\left(4x^3+3x^2+2x+1\right)}{\left(x-1\right)^2\left(2x+4\right)}=Lim_{x\rightarrow1}\frac{4x^3+3x^2+2x+1}{2x+4}=\frac{5}{3}\)
Vậy \(Lim_{x\rightarrow1}\frac{4x^5-5x^4+1}{\left(x-1\right)\left(x^3+x-2\right)}=\frac{5}{3}\)
bài này tớ trình bày ra cho dễ hiểu , nếu làm thì bỏ qua bước cơ bản là được
\(Lim_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=Lim_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=Lim_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{1+1+1+1+1}{1+1+1}=\frac{5}{3}\)
Vậy \(Lim_{x\rightarrow1}\frac{x^5+1}{x^3+1}=\frac{5}{3}\)
